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–
Gordon♦Feb 1 '11 at 11:19

First we need a day from that week so by knowing the week number and knowing that a week has seven days we are going to do so the

$pickADay = ($weekNo-1) * 7 + 3;

this way pickAday will be a day in our desired week.

Now because we know the year we can check which day is that.
things are simple if we only need dates newer than unix timestamp
We will get the unix timestamp for the first day of the year and add to that 24*3600*$pickADay and all is simple from here because we have it's timestamp we can know what day of the week it is and calculate the head and tail of that week accordingly.

If we want to find out the same thing of let's say 12th week of 1848 we must use another approach as we can not get the timestamp. Knowing that each year a day advances 1 weekday meaning (1st of november last year was on a sunday, this year is on a monday, exception for the leap years when it advances 2 days I believe, you can check that ). What I would do if the year is older than 1970 than make a difference between it and the needed year to know how many years are there, calculate the day of the week as my pickADay was part of 1970, shift it back one weekday for each. $shiftTimes = ($yearDifference + $numberOfLeapYears)%7, in the difference. shift the day backwords $shiftTimes, then you will know what day of the week was that day those years ago, then find the weekhead and weektail. Same thing can be used also for the future if it seems simpler. Try it if it works and tell me if it does not.

function week_date($week, $year){
$date = new DateTime();
return "first day of the week is ".$date->setISODate($year, $week, "1")->format('Y-m-d')
."and last day of the week is ".$date->setISODate($year, $week, "7")->format('Y-m-d');
}
echo week_date(12,2014);